If the system of equations x + 4 y - z =, 7 x + 9 y + z = - 3 , 5 x + y + 2 z = - 1 has infinitely many solutions, then ( 2+ 3 ) is equal to: [2024]

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Solution

Q.
If the system of equations $x + 4 y - z = λ,$ $7 x + 9 y + μ z = - 3,$ $5 x + y + 2 z = - 1$ has infinitely many solutions, then $(2 μ + 3 λ)$ is equal to: [2024]

1 $- 2$

2 $3$

3 $- 3$

4 $2$

Ans.
(3)

We have, $x + 4 y - z = λ$

$7 x + 9 y + μ z = - 3; 5 x + y + 2 z = - 1$

$[\begin{matrix} 1 & 4 & - 1 \\ 7 & 9 & μ \\ 5 & 1 & 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} λ \\ - 3 \\ - 1 \end{matrix}]$

$A = [\begin{matrix} 1 & 4 & - 1 \\ 7 & 9 & μ \\ 5 & 1 & 2 \end{matrix}], B = [\begin{matrix} λ \\ - 3 \\ - 1 \end{matrix}]$

$A X = B \Rightarrow X = A^{- 1} B = \frac{adj A}{| A |} B$

Since, the system has infinitely many solutions,

$∴$ $| A | = 0 and (adj A) \cdot B = 0 .$

Now, $| A |$ $= 0 \Rightarrow 18 - μ - 4 (14 - 5 μ) - 1 (7 - 45) = 0$

$\Rightarrow 18 - μ - 56 + 20 μ + 38 = 0$

$\Rightarrow 19 μ = 0 \Rightarrow μ = 0$

Also, $adj A = [\begin{matrix} 18 & - 9 & 9 \\ - 14 & 7 & - 7 \\ - 38 & 19 & - 19 \end{matrix}]$

$(adj A \cdot B) = 0$

$\Rightarrow [\begin{matrix} 18 & - 9 & 9 \\ - 14 & 7 & - 7 \\ - 38 & 19 & - 19 \end{matrix}] [\begin{matrix} λ \\ - 3 \\ - 1 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

$18 λ + 27 - 9 = 0 \Rightarrow 18 λ = - 18 \Rightarrow λ = - 1$

Hence, $2 μ + 3 λ = 2 (0) + 3 (- 1) = - 3$

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